Answer:
Your options are not clear
Step-by-step explanation:
[tex]\sqrt[3]{-1000 \times p^{12} \times q^3} \\\\(-1 \times 10^3 \times p^{12} \times q^3)^{\frac{1}{3} }\\\\(-1^3)^{\frac{1}{3} }\times 10^{3 \times \frac{1}{3} } \times p^{12 \times \frac{1}{3}} \times q^{3 \times \frac{1}{3}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ (-1)^3 = - 1 \ ] \\\\- 1 \times 10 \times p^4 \times q\\\\-10p^4q[/tex]
Express the following repeating decimal as a fraction in simplest form.
Answer:
[tex]0.\overline{369} = \frac{41}{111}[/tex]
Step-by-step explanation:
x = 0.369369369...
10x = 3.69369369...
100x = 36.9369369...
1000x = 369.369369...
1000x - x = 369
999x = 369
[tex]x = \frac{369}{999} \\\\x = \frac{123}{333}\\\\x = \frac{41}{111}[/tex]
Given that yx is a diameter of V, find mZVX if mYVZ=(15r -1)", mWVX =(8x- 2).mUVW =(7x +15).
and mYVU = 3x + 5).
A 32
B 45°
C 70
D. 134
Find the value of y and show work
Answer:
75
Step-by-step explanation:
∠K and ∠ R are congruent (equal)
Triangle Sum Theory - angles of all triangles add to 180
180 - 79 - 26 = 75
Segment [tex]$s_1$[/tex] has endpoints at [tex]$(3+\sqrt{2},5)$[/tex] and[tex]$(4,7)$[/tex]. Segment [tex]$s_2$[/tex] has endpoints at [tex]$(6-\sqrt{2},3)$[/tex] and[tex]$(3,5)$[/tex]. Find the midpoint of the segment with endpoints at the midpoints of [tex]$s_1$[/tex] and [tex]$s_2$[/tex]. Express your answer as [tex]$(a,b)$[/tex].
Answer:
The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).
Step-by-step explanation:
Midpoint of a segment:
The coordinates of the midpoint of a segment are the mean of the coordinates of the endpoints of the segment.
Midpoint of s1:
Using the endpoints given in the exercise.
[tex]x = \frac{3 + \sqrt{2} + 4}{2} = \frac{7 + \sqrt{2}}{2}[/tex]
[tex]y = \frac{5 + 7}{2} = \frac{12}{2} = 6[/tex]
Thus:
[tex]M_{s1} = (\frac{7 + \sqrt{2}}{2},6)[/tex]
Midpoint of s2:
[tex]x = \frac{6 - \sqrt{2} + 3}{2} = \frac{9 - \sqrt{2}}{2}[/tex]
[tex]y = \frac{3 + 5}{2} = \frac{8}{2} = 4[/tex]
Thus:
[tex]M_{s2} = (\frac{9 - \sqrt{2}}{2}, 4)[/tex]
Find the midpoint of the segment with endpoints at the midpoints of s1 and s2.
Now the midpoint of the segment with endpoints [tex]M_{s1}[/tex] and [tex]M_{s2}[/tex]. So
[tex]x = \frac{\frac{7 + \sqrt{2}}{2} + \frac{9 - \sqrt{2}}{2}}{2} = \frac{16}{4} = 4[/tex]
[tex]y = \frac{6 + 4}{2} = \frac{10}{2} = 5[/tex]
The midpoint of the segment with endpoints at the midpoints of s1 and s2 is (4,5).
21(2-y)+12y=44 find y
Answer:
[tex]\textbf{HELLO!!}[/tex]
[tex]21\left(2-y\right)+12y=44[/tex]
[tex]42-21y+12y=44[/tex]
[tex]~add ~similar\:elements[/tex]
[tex]42-9y=44[/tex]
[tex]Subtract~42~from~both~sides[/tex]
[tex]42-9y-42=44-42[/tex]
[tex]-9y=2[/tex]
[tex]Divide\:both\:sides\:by\:}-9[/tex]
[tex]\frac{-9y}{-9}=\frac{2}{-9}[/tex]
[tex]y=-\frac{2}{9}[/tex]
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hope it helps...
have a great day!
When we expand (2x + 1/2)^6, what is the coefficient on the x^4 term?
Answer: The coefficient before x^4 is 60
Step-by-step explanation:
Hey! So I am not an expert at this, but you have to use the binomial theorem
I have attached of the Pascals Triangle (one shows the row numbering as well)
Basically in a pascal triangle, you add the two numbers above it to get the next number below
As you can see, the rows start from 0 instead of 1
The 6th row contains the numbers 1, 6, 15, 20, 15, 6, 1 which would be the coefficient terms
NOTE: the exponents always add to 6, the first term starts at 6 and decrease it's exponent by 1 each time (6, 5, 4, 3, 2, 1, 0) and the second term increases it's exponent by 1 each time (0, 1, 2, 3, 4, 5, 6)
Using this information the third term from the sixth row (15) would be where it is x^4 (I have circled it on the second image)
It would be 15 × 2^4 × (1/2)^2 = 60
The reason why it is 2^4 and (1/2)^2 is because the third term has the exponents 4 and 2 (bolded on the NOTE) which means that the first term must be put to the power of 4 and the second term must be put to the 2nd power
Sorry for the lousy explanation. I really hope this makes sense! Let me know if this helped :)
What is the solution to the inequality x(x – 3) > 0?
Answer:
The solution to the inequality is [tex](-\infty, 0) \cup (3, \infty)[/tex]
Step-by-step explanation:
We have a product, which is positive if both terms is positive or if both is negative.
Both positive:
[tex]x > 0[/tex]
[tex]x - 3 > 0 \rightarrow x > 3[/tex]
Then the intersection of these two is: [tex]x > 3[/tex]
Both negative:
[tex]x < 0[/tex]
[tex]x - 3 < 0 \rightarrow x < 3[/tex]
Then the intersection of those two is: [tex]x < 0[/tex]
Then:
Union of two solutions:
[tex]x < 0[/tex] or [tex]x > 3[/tex]
Then
[tex](-\infty, 0) \cup (3, \infty)[/tex]
Simplify the expression
Answer:
6
Step-by-step explanation:
3 sqrt(20) / sqrt(5)
We know that sqrt(a) /sqrt(b) = sqrt(a/b)
3 sqrt(20/5)
3 sqrt(4)
3 *2
6
Simplify -4 + (-3) + 6.
Answer:3/6 in simplest fraction form is 1/2.
Step-by-step explanation:EASY and my chanel is FireFlameZero if u can check dat out
Rewrite the equation by completing the square.
x^2 + 7x + 12 = 0
Answer:
x^2 + 7x + 12 = 0
x^2 + 7x = -12
(+3)(+4)=0
=−3
=−4
I also love r o blox
Hope This Helps!!!
Answer:
(x + [tex]\frac{7}{2}[/tex] )² - [tex]\frac{1}{4}[/tex] = 0
Step-by-step explanation:
Given
x² + 7x + 12 = 0
To complete the square
add/subtract ( half the coefficient of the x- term)² to x² + 7x
x² + 2([tex]\frac{7}{2}[/tex] )x + [tex]\frac{49}{4}[/tex] - [tex]\frac{49}{4}[/tex] + 12 = 0
(x + [tex]\frac{7}{2}[/tex] )² - [tex]\frac{49}{4}[/tex] + [tex]\frac{48}{4}[/tex] = 0 , that is
(x + [tex]\frac{7}{2}[/tex] )² - [tex]\frac{1}{4}[/tex] = 0
The following data represents the number of days absent and the final grade for a sample of college students in a general education course at a large state university.
No. of absences 0 1 2 3 4 5 6 7 8 9
Final Grade 89.2 86.4 83.5 81.1 78.2 73.9 64.3 71.8 65.5 66.2
a) Which variable is the explanatory variable?
b) Draw a scatter plot and describe your scatter plot (Direction, Strength, Form).
c) Compute the correlation coefficient
d) Does a linear relation exist between the number of absences and the final grade? Justify your answer.
e) Write out the least-squares regression line equation.
f) Compute and draw the residual (on your scatter plot) for a student who misses 5 class meetings.
g) Explain the slope in context.
h) Is the y-intercept meaningful in this situation? Explain.
i) Compute and interpret the coefficient of determination.
j) Construct a residual plot to verify the requirements of the least-squares regression model.
